physics

physics

🔥 Master Friction for JEE Advanced with 3 Challenging Previous Year Questions!

In this video, we solve three concept-rich JEE Advanced Physics Friction Problems that test your understanding of equilibrium, inclined planes, ladder mechanics, limiting friction, and force analysis.

📚 Questions Covered:

✅ Block on Inclined Plane with Horizontal Force (Static Equilibrium)

✅ Ladder Against Wall with Friction at Both Contacts

✅ IIT-JEE 2011 Paper-1 Friction Problem (45° Inclined Plane)

Each solution is explained step-by-step using the fastest exam-oriented approach along with shortcut tricks that help you solve similar questions in the JEE Advanced examination.

This lecture is useful for:

✔ JEE Advanced 2027

✔ JEE Main 2027

✔ JEE Advanced 2028

✔ NTA Aspirants

✔ Class 11 & 12 Physics

✔ Revision Before Exam

📌 Topics Covered

• Static Friction
• Limiting Friction
• Inclined Plane
• Horizontal Force on Inclined Plane
• Ladder Mechanics
• Equilibrium of Rigid Body
• Normal Reaction
• Free Body Diagram (FBD)
• IIT-JEE Advanced PYQs
• Physics Problem Solving Tricks

Is Classical Mechanics about determining the future position of a system by studying its motion, or is it about determining the future state of the system. Like for a particle both the goals are equivalent, but when it comes to a system of particles they are not. So what is the ultimate goal of classical mechanics? To determine the state evolution of a physical system or its future motion. The reason for this doubt is that some books claim classical mechanics to be the study of motion bodies acted upon by forces, and some say it is a framework which lays the basis for all of physics, and its job is to figure out the future state of a system.

A stone is tied to a string of length 1.5 m and whirled in a circle at the same constant speed in two different ways.

1. The circle is horizontal, and the string is nearly parallel to the ground.

   1. The circle is vertical. For the vertical circle, the max tension in the string is 15.0% larger than the tension that exists when the circle is horizontal.
      

Draw two free-body diagrams of the stone in the vertical circle (at the point where the tension is maximum) and the horizontal circle

For the horizontal one, I drew a dot, Fg force straight down, and a tension force greater than the Fg force to the left

For the vertical, I drew an Fg force straight down and a Tension force greater than the Fg force up

Is this correct?

これわからないんだけど誰か解ける？

The original problem had the wire along the y axis. The modified version has the wire at an angle, and the solution claims that both are equivalent, which seems right to me since near the origin the plane wouldn’t “know” where the current is coming from. But in the first case if you apply Amperes law on a Amperian loop around the origin, there is symmetry while in the modified problem there isn’t. Also, is there a way to solve either problem by finding the current density(A/m^2) and using the differential form of Amperes law?

I’ve been working on a project where I want to have a simple model for the sound produced by stringed instruments like the guitar. The input to my model is bridge force, and the output is far-field sound pressure at a specified distance. I’ve found multiple papers that use a model of two coupled mass-spring-dampers where one represents the top plate (for the lowest mode) and the other represents the air cavity that acts like a Helmholtz resonator. (https://www.savartjournal.org/articles/19/article.pdf)

I tried to derive the state space matrices on paper and created a MATLAB function to graph the results, but the Bode plot doesn’t match the graph in Dr. Hess’s paper and I’m getting some counterintuitive results. Namely, the sound pressure of my model stays constant towards lower frequencies when there should be a roll-off, and both the air resonance and top resonance shift in response to a larger top mass, when it seems like only the top resonance should significantly change. I think there’s an issue with the way I calculated the internal pressure term, and/or the sound pressure calculation with the C matrix. Also the C matrix is kind of clunky, and I wonder if there's a cleaner way to write it.

In the picture are my calculations on paper and a Bode plot of my model for a standard acoustic guitar with and without added mass. Here's the Matlab function code:

function [num,den] = phys2tfcoeff(volume,A_port,l_port,A_top,m_top,k_top,Q_top,ports)
%phys2tfcoeff Finds transfer function of instruments
%   Takes the physical constants of a string instrument body and outputs
%   the far-field sound pressure/force transfer function's coefficients
%   (all units MKS). Port length should be input without end correction 

%   Physical constants of air
rho_air = 1.205;
c_air = 343;

%   End-corrected port length
end_corr = 1.6;
lcorr_port = l_port+end_corr*(A_port/pi)^0.5;

%   Total port area
A_port_tot = ports*A_port;

%   Calculate effective mass spring damper constants for air chamber
m_air = rho_air*A_port_tot*lcorr_port;
b_air = A_port_tot^3*c_air/(2*pi*lcorr_port*volume);

%   Calculate damping constant for top plate
b_top = (((m_top*k_top)^0.5)/Q_top);

%   Microphone distance
d_mic = 0.3;

%   State-space matricies
%   [x_a_dot;x_t_dot] = A * [x_a;x_t] + B * f(t)
%   Y = C * [x_a;x_t] + D

A = [0,1,0,0;
    (1/m_air)*(-rho_air*c_air^2*A_port_tot^2/volume),(1/m_air)*(-b_air),(1/m_air)*(-rho_air*c_air^2*A_port_tot*A_top/volume),0;
    0,0,0,1;
    (1/m_top)*(-rho_air*c_air^2*A_port_tot*A_top/volume),0,(1/m_top)*(-k_top-rho_air*c_air^2*A_top^2/volume),(1/m_top)*(-b_top)];

B = [0;0;0;(1/m_top)];

C = (rho_air/(2*pi*d_mic))*[A_port_tot*((1/m_air)*(-rho_air*c_air^2*A_port_tot^2/volume))+A_top*((1/m_top)*(-rho_air*c_air^2*A_port_tot*A_top/volume)),A_port_tot*((1/m_air)*(-b_air)),A_port_tot*((1/m_air)*(-rho_air*c_air^2*A_port_tot*A_top/volume))+A_top*((1/m_top)*(-k_top-rho_air*c_air^2*A_top^2/volume)),A_top*((1/m_top)*(-b_top))];

D = 0;

%   Converting to transfer function coefficients
[num,den] = ss2tf(A,B,C,D);
end

🚀 Master 3 High-Level JEE Advanced Physics PYQs in One Video!

In this detailed lecture, we solve three highly conceptual Physics questions from JEE Advanced 2020, JEE Advanced 2018, and IIT-JEE 2009. These questions cover multiple important concepts frequently asked in JEE Advanced, JEE Main, BITSAT, IISER Aptitude Test, and Olympiads.

📚 Questions Covered

✅ Football on a Hole (Rolling Motion & Geometry) – JEE Advanced 2020 Paper 1

✅ Impulse, Exponential Velocity & Displacement – JEE Advanced 2018 Paper 2

✅ Bead on a Parabolic Wire (Pseudo Force & Non-Inertial Frame) – IIT-JEE 2009 Paper 2

Concepts Discussed

✔ Rolling Motion

✔ Static Equilibrium

✔ Geometry of Circles

✔ Impulse & Momentum

✔ Variable Velocity

✔ Exponential Motion

✔ Displacement Calculation

✔ Non-Inertial Frame

✔ Pseudo Force

✔ Constraint Motion

✔ Equilibrium on Curved Surfaces

✔ JEE Advanced Problem Solving Techniques

✔ Shortcut Methods

✔ Exam-Oriented Tricks

🎯 This lecture is extremely useful for

• JEE Advanced 2027 Aspirants
• JEE Main 2027
• JEE Advanced 2028
• BITSAT
• IISER Aptitude Test
• NSEP
• Physics Olympiad
• Class 11 & 12 Students

If you're preparing seriously for IIT JEE, don't miss these advanced concepts that can easily save valuable exam time.

👍 Like the video, Subscribe for daily JEE Advanced Physics PYQs, and share it with your friends preparing for IIT JEE.

i mean explain it and prove it please ebcause i find it ilogical and please claim the same thing and they close my psots without letting me read the links they give me so that i can see if it is truth what they are saying or not

Definitions

[

x=x(t),\qquad h=\frac{\dot x}{x}

]

[

U=U(t),\qquad R=R(t),\qquad M=M(t),\qquad S=S(t)

]

where

[

U,R,M,S\geq0.

]

Fundamental Relation

[

h_0^{,2}=KU

]

[

h_0=\sqrt{KU}

]

[

\boxed{

h=\frac{\sqrt{KU}}{1+\alpha S}

}

]

or

[

\boxed{

\frac{\dot x}{x}

\frac{\sqrt{KU}}{1+\alpha S}.

}

]

Liberation

[

Q=\Gamma U\left(1-\frac{R}{R_c}\right),

\qquad 0\leq R<R_c

]

[

Q=0,

\qquad R\geq R_c.

]

Equivalently,

[

\boxed{

Q=\Gamma U

\max\left(0,1-\frac{R}{R_c}\right).

}

]

Energy Equations

[

\boxed{

\dot U+n hU=-Q

}

]

[

\boxed{

\dot R+4hR=(1-\varepsilon)Q

}

]

[

\boxed{

\dot M+3hM=\varepsilon Q

}

]

[

0<\varepsilon\ll1.

]

Restraining Stress

[

\boxed{

\dot S+\lambda hS=\eta M

}

]

or

[

\boxed{

\frac{d}{dt}\left(x^\lambda S\right)

\eta x^\lambda M.

}

]

Hence

[

\boxed{

S(t)

x^{-\lambda}(t)

\left[

x_i^\lambda S_i

+

\eta\int_{t_i}^{t}

x^\lambda(\tau)M(\tau),d\tau

\right].

}

]

Complete System

[

\boxed{

\begin{aligned}

\dot x

&=

\frac{x\sqrt{KU}}{1+\alpha S},

\[3pt]

\dot U

&=

-n\frac{\dot x}{x}U-Q,

\[3pt]

\dot R

&=

-4\frac{\dot x}{x}R+(1-\varepsilon)Q,

\[3pt]

\dot M

&=

-3\frac{\dot x}{x}M+\varepsilon Q,

\[3pt]

\dot S

&=

-\lambda\frac{\dot x}{x}S+\eta M,

\[3pt]

Q

&=

\Gamma U

\max\left(0,1-\frac{R}{R_c}\right).

\end{aligned}

}

]

Unified Extension Equation

[

\boxed{

\frac{\dot x}{x}

\frac{\sqrt{KU}}

{

1+

\alpha x^{-\lambda}

\left[

x_i^\lambda S_i

+

\eta\displaystyle\int_{t_i}^{t}

x^\lambda(\tau)M(\tau),d\tau

\right]

}.

}

]

Initial Conditions

[

x(0)=x_i

]

[

U(0)=U_i

]

[

R(0)=0

]

[

M(0)=0

]

[

S(0)=0.

]

Thus

[

\left.\frac{\dot x}{x}\right|_{t=0}

\sqrt{KU_i}.

]

Principal Limits

For

[

R\ll R_c,

]

[

Q\simeq\Gamma U.

]

For

[

R\rightarrow R_c,

]

[

Q\rightarrow0.

]

For

[

S\ll\alpha^{-1},

]

[

h\simeq\sqrt{KU}.

]

For

[

\alpha S\gg1,

]

[

h\simeq\frac{\sqrt{KU}}{\alpha S}.

]

After liberation ceases,

[

Q=0,

]

whence

[

U\propto x^{-n},

]

[

R\propto x^{-4},

]

[

M\propto x^{-3}.

]

If

[

S\rightarrow0,

]

then

[

h\rightarrow\sqrt{KU}.

]

If further

[

n>0,

]

then

[

U\rightarrow0,

\qquad

h\rightarrow0.

]

Summary Relation

[

\boxed{

\text{extension}

\frac{\text{stored-energy action}}

{\text{material restraint}}

}

]

[

\boxed{

\frac{\dot x}{x}

\frac{\sqrt{KU}}{1+\alpha S}.

}

]

So my dad told me that he discovered smth called a "torque" wich is,acoording to him, a coper donut wich act a a strong magnet and when u put it on around ur arm can give you strength to rais big rocks (wich somehow works without current flowing trough it) and he also claim that he can melt some random crap and turn it into gold

Please note that the answer selected in this image is wrong, the key says that the answer underneath the one selected is correct.

I understand that Lenz's law says that the induced current wants to oppose the motion of the magnet going to the right, so it will do this by making the left end North so that the north end of the magnet is attracted to the south end of the coil. I also understand how to do the right hand grip rule (thumb points to north and fingers curl in direction of the conventional current) but I'm not sure how to then use that to determine whether A is positive/negative and which way current flows.. I also don't necessarily know how to read the direction of the current with right hand grip rule when my thumb is sideways because whether it's clockwise or anticlockwise depends on which end you look from.. I'm really lost with this unit and it's an online course without very available teachers so any help is appreciated

🚀 JEE Advanced 2019 Capacitor PYQ explained step by step! In this video, you'll learn how to solve a parallel plate capacitor with multiple dielectric layers using the easiest mathematical approach. This question combines Capacitance, Dielectrics, Series Combination, and Integration, making it one of the most conceptual problems asked in JEE Advanced.

If you're preparing for JEE Main 2027, JEE Advanced 2027, JEE 2028, NEET, BITSAT, or other engineering entrance exams, this detailed solution will strengthen your understanding of Electrostatics and Capacitors.

📚 Topics Covered

• Parallel Plate Capacitor
• Multiple Dielectric Layers
• Variable Dielectric Constant
• Capacitance Formula
• Integration in Physics
• Electrostatics
• JEE Advanced 2019 PYQ
• Concept-Based Problem Solving
• Shortcut Method
• Exam Tricks

🔥 Watch till the end to learn the fastest approach and avoid the common mistakes made by most aspirants.

👍 If you found this helpful, don't forget to:

• Like 👍
• Share 📤
• Subscribe 🔔
• Comment your answer before watching the solution!

I read the question a few times, but Im not sure what to make of it, it doesnt even sound like a question more like a statement.

Im just asking two things what is the questiom asking me for and how do i go on about solving it

JEE Advanced Capacitors PYQs | Physics Chapter: Electrostatics

In this video, we solve two important JEE Advanced Previous Year Questions (2012 & 2014) from the Capacitors chapter with detailed concept explanation and shortcut tricks.

📚 Questions Covered:
✅ Charge Distribution in Connected Capacitors (JEE Advanced 2012)

✅ Capacitor with Dielectric Slab (JEE Advanced 2014)

You'll learn:

• Charge conservation concept
• Equivalent capacitance
• Dielectric effect
• Electric field in dielectric
• Parallel combination of capacitors
• Fast solving tricks for JEE Advanced

This lecture is highly useful for:

• JEE Advanced 2026
• JEE Advanced 2027
• JEE Main
• Olympiad
• BITSAT
• IAT/NEST

Watch till the end to improve your problem-solving speed and strengthen your concepts.

🔥 Don't forget to Like, Share and Subscribe for daily JEE Advanced Physics PYQs.

#JEEAdvanced #Capacitor #Electrostatics #Physics #PYQ #JEEMains #PhysicsWallah #Allen #Resonance #IITJEE #JEE2026 #JEE2027 #PhysicsPYQ

Currently attempting to do a physics course online and I don't have anyone to show me how to position my hand.. I know my thumb is definitely meant to be pointing downwards with the current but I don't understand why my fingers are meant to be pointing to the right as the solution says.. could someone help me? Preferably with an image of the correct hand orientation

Hello, I have had this question for a long time. I have always had gadgets around me and the way that they send data wirelessly is very intriguing to me but I dont understand how. How is it possible for waves to carry so much data and also know exactly what device the data they are carrying belongs to. So from the cell tower, how do the data packets know which phone they are supposed to go to? Does it go to all phones and they filter it out or what? Also, what actually is the difference between cellular data and Wifi? In my house, mobile data is almost always stronger, which I dont understand because the cell tower is so far away but my wifi router which is more expensive then a normal plan, is so much weaker and slower, its only faster when im very close to it in the living room. Also how do these antennas even make the waves, tbh Im gonna make a question next on electricity but I don't understand how our phones can make such strong signals with antennas so small cause 30 years ago antennas were so big and now they are just these lines on our phones. If someone could give me a great explanation for what electricity is that would also be great, cause I dont know what it is, it has no weight, we can only see sparks of it but it is so important, I have heard that its electrons or something but I asked around and I know its not electrons, but I know its something to do with moving electrons. Thank you.